Practice <strong>NEET UG Physics</strong> MCQ questions covering Mechanics, Thermodynamics, Waves & Optics, Modern Physics, Electricity & Magnetism, and Electromagnetic Induction. Aligned with NCERT Class 11 & 12 Physics syllabus.
A body is moving with a constant velocity of 10 m/s. What is the net force acting on it?
Answer: A
According to Newton's first law, when velocity is constant, acceleration is zero. Therefore, net force F = ma = m × 0 = 0 N.
Q.2Easy
If the mass of an object is doubled and the force applied remains the same, how does the acceleration change?
Answer: B
From Newton's second law, F = ma. If F is constant and m doubles, then a = F/2m, so acceleration becomes half.
Q.3Easy
A 5 kg object experiences a net force of 20 N. What is its acceleration?
Answer: B
Using F = ma, we get a = F/m = 520 = 4 m/s².
Q.4Easy
Which of the following is NOT a consequence of Newton's third law of motion?
Answer: B
Newton's third law states action and reaction act on different bodies, not the same body. They are equal in magnitude, opposite in direction, and act simultaneously.
Q.5Medium
A block of mass m slides down a frictionless incline at angle θ. What is the acceleration along the incline?
Answer: A
The component of gravitational force along the incline is mg sin θ. Using F = ma, we get a = g sin θ.
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Q.6Medium
Two blocks of masses m₁ = 3 kg and m₂ = 2 kg are connected by a light string and pulled horizontally with force F = 10 N. What is the tension in the string between them?
Answer: B
Total acceleration a = F/(m₁+m₂) = 510 = 2 m/s². For block m₂, tension T = m₂ × a = 2 × 2 = 4 N.
Q.7Medium
A car of mass 1000 kg accelerates uniformly from rest to 20 m/s in 10 seconds on a horizontal road. If friction force is 500 N, what is the driving force?
Answer: C
Acceleration a = (20-0)/10 = 2 m/s². Net force = ma = 1000 × 2 = 2000 N. Driving force = Net force + friction = 2000 + 500 = 2500 N.
Q.8Medium
A 50 kg person stands on a weighing machine in an elevator. When the elevator accelerates downward at 2 m/s², what is the reading on the scale? (g = 10 m/s²)
Answer: A
Apparent weight = m(g - a) = 50(10 - 2) = 50 × 8 = 400 N.
Q.9Easy
Three forces of 10 N, 20 N, and 30 N act on a body in the same direction. What is the resultant force?
Answer: D
When forces act in the same direction, the resultant is the sum of all forces: 10 + 20 + 30 = 60 N.
Q.10Medium
A 2 kg object moving at 5 m/s collides with a wall and comes to rest in 0.1 seconds. What is the average force exerted by the wall?
Answer: B
Acceleration a = (v-u)/t = (0-5)/0.1 = -50 m/s². Force F = ma = 2 × (-50) = -100 N. Magnitude = 100 N.
Q.11Easy
A book rests on a table. The normal force exerted by the table on the book is an example of:
Answer: C
Normal force is a contact force arising from physical contact between the table and book. The action-reaction pair would be the weight of the book and the force the book exerts on the table.
Q.12Medium
A body moving in a circle with constant speed experiences acceleration. This acceleration is directed:
Answer: C
In uniform circular motion, acceleration (centripetal) is always directed toward the center of the circle (radially inward).
Q.13Medium
A 100 N force is applied on a 10 kg block on a horizontal surface with kinetic friction coefficient μₖ = 0.2. What is the acceleration? (g = 10 m/s²)
Answer: A
Normal force N = mg = 10 × 10 = 100 N. Friction force f = μₖN = 0.2 × 100 = 20 N. Net force = 100 - 20 = 80 N. Acceleration = 1080 = 8 m/s².
Q.14Medium
An object of mass 5 kg is suspended by a rope. The rope can withstand a maximum tension of 60 N. If the object is accelerated upward, what is the maximum upward acceleration? (g = 10 m/s²)
Answer: A
Tension T = m(g + a). Maximum T = 60 N. So 60 = 5(10 + a), which gives 12 = 10 + a, therefore a = 2 m/s².
Q.15Medium
Two objects A and B have the same mass but A is moving at 10 m/s and B is moving at 5 m/s. If equal forces bring both to rest, which one travels a greater distance?
Answer: A
Using v² = u² + 2as, with same deceleration, the object with higher initial velocity (A) travels a greater distance.
Q.16Medium
A 2 kg mass and a 3 kg mass are connected by a light string over a frictionless pulley. When released, what is the acceleration of the system? (g = 10 m/s²)
A stone is thrown vertically upward from the ground. At the highest point, the velocity is zero but acceleration is not. What is the acceleration at the highest point?
Answer: B
At the highest point, velocity is zero, but the gravitational force continues to act, producing an acceleration of g = 10 m/s² downward.
Q.18Easy
A car moving at 72 km/h brakes with a deceleration of 5 m/s². How long does it take to stop?
Answer: B
72 km/h = 20 m/s. Using v = u - at, 0 = 20 - 5t, so t = 4 seconds.
Q.19Hard
A 6 kg block is placed on a 10 kg block on a frictionless surface. The coefficient of friction between the blocks is 0.3. If a horizontal force of 32 N is applied to the 10 kg block, what is the acceleration of the 6 kg block? (g = 10 m/s²)
Answer: C
Maximum static friction on 6 kg block = μ × m × g = 0.3 × 6 × 10 = 18 N. If blocks move together: a = 32/(6+10) = 2 m/s². Force needed for 6 kg block = 6 × 2 = 12 N < 18 N. So blocks move together with a = 2 m/s². But checking: friction available = 18 N can support up to a = 618 = 3 m/s². System acceleration = 1632 = 2 m/s². The 6 kg block needs 12 N which is available, so it moves at 2 m/s². Rechecking: If 10 kg block alone: a₁₀ = 1032 = 3.2 m/s². Max acceleration for 6 kg = 618 = 3 m/s². They slip. 6 kg block accelerates at 618 = 3 m/s².
Q.20Medium
A projectile is launched at an angle of 45° with initial velocity 20 m/s. At the highest point of trajectory, what is the magnitude of acceleration? (g = 10 m/s²)
Answer: C
At any point in projectile motion including the highest point, only gravitational acceleration acts, which is g = 10 m/s² directed downward.
